Question 1049827

let one number be {{{x}}} and another {{{y}}}

if one number is {{{4}}} more than another, means 

{{{x=y+4}}} ........eq.1

if the difference between their squares is {{{104}}}, means 

{{{x^2-y^2=104}}} ........eq.2

substitute {{{x}}} from eq.1 in eq.2

{{{(y+4)^2-y^2=104}}} ........eq.2......solve for {{{y}}}

{{{y^2+8y+16-y^2=104}}} 

{{{cross(y^2)+8y+16-cross(y^2)=104}}} 

{{{8y=104-16}}} 

{{{8y=88}}} 

{{{y=88/8}}} 

{{{y=11}}} 

now find {{{x}}}


{{{x=y+4}}} ........eq.1
{{{x=11+4}}} 
{{{x=15}}} 

so, one number is {{{15}}} and another {{{11}}}